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Nina [5.8K]
3 years ago
7

One of the pieces of evidence supporting energy quantization was the line spectra of elements. Why does this demonstrate energy

quantization?
a) Photons are packets of energy that act both as a particle and a wave.
b) The spectra is continuous and shows emission at all wavelengths.
c) There are sharp emission lines demonstrating discrete energy levels.
d) All compounds emit light in the ultraviolet and visible regions specifically.​
Chemistry
1 answer:
bezimeni [28]3 years ago
8 0

Answer:

c) There are sharp emission lines demonstrating discrete energy levels.

Explanation:

When an element emits energy in the form of radiation, it produces a spectrum of colors on a photographic plate. This spectrum can either be continuous or discrete. In continuous spectrum the spectrum continues without any discrimination between two regions. This represents the continuous emission of radiation, and thus the continuous emission of energy without any break.

On the other hand, the line spectrum consists of discrete and sharp lines, which shows the emission of radiation in a certain amount in a certain time, with a break between emission. Hence, the line spectra supports the quantization of energy.

The correct option is:

<u>c) There are sharp emission lines demonstrating discrete energy levels.</u>

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_____ bonds are formed when the electronegativity difference is 0.4 &gt; 2.0. Give two examples:
34kurt

<u>Ionic Bond</u> is formed when the electronegativity difference is 0.4 > 2.0. Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.

Explanation:

Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.

An  electronegativity of an atom is affected by

  • The atomic number of the atom
  • Secondly by the distance at which the valence electron are residing from the nucleus

1. In case the  electronegativity difference (which is denoted by  ΔEN) is less than 0.5  then the bond formed is known as N<u>onpolar covalent. </u>

2. In case  the ΔEN is in between 0.5 and 1.6, the bond formed is referred to as the<u> Polar covalent </u>

3. In case  the ΔEN is more /greater  than 2.0, then the bond formed is referred to as<u> Ionic Bond</u>

<u>2 Examples of Ionic bonds</u>

  • The formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom is an example of Ionic bond formation.
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6 0
3 years ago
The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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3 years ago
Calculate the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 ml solution.
Lelechka [254]

Taking into account the definition of molarity, the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 mL solution is 2 \frac{moles}{liter}.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

molarity=\frac{number of moles}{volume}

Molarity is expressed in units \frac{moles}{liter}.

<h3>Molarity in this case</h3>

In this case, you have:

  • number of moles= 0.2 moles
  • volume= 100 mL= 0.1 L

Replacing in the definition of molarity:

molarity=\frac{0.2 mole}{0.1 L}

Solving:

<u><em>molarity= 2 </em></u>\frac{moles}{liter}

Finally, the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 mL solution is 2 \frac{moles}{liter}.

Learn more about molarity:

brainly.com/question/9324116

brainly.com/question/10608366

brainly.com/question/7429224

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